Find $\dfrac{d}{dx}\left([\ln(x)]^3\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $3[\ln(x)]^2$ (Choice B) B $3\left(\dfrac1x\right)^2$ (Choice C) C $\left(\dfrac1x\right)^3$ (Choice D) D $\dfrac{3[\ln(x)]^2}{x}$
Explanation: $[\ln(x)]^3$ is a composition of two, more basic, functions: $\ln(x)$ and $x^3$. In other words, suppose $u(x)=\ln(x)$ and $v(x)=x^3$, then $[\ln(x)]^3=v\Bigl(u(x)\Bigr)$, or $(v\circ u)(x)$. Therefore, $\dfrac{d}{dx}\left([\ln(x)]^3\right)$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=x^3$, and therefore $v'(x)=3x^2$. Now we plug $u(x)=\ln(x)$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\left(\ln(x)\right) \\\\ &={3[\ln(x)]^2} \end{aligned}$ Finding $u'(x)$ $u(x)=\ln(x)$, and therefore $u'(x)={\dfrac1x}$. Putting things together $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left([\ln(x)]^3)\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=\ln(x)\text{, }v(x)=x^3} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={3[\ln(x)]^2}\cdot {\dfrac1x} \\\\ &=\dfrac{3[\ln(x)]^2}{x} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left([\ln(x)]^3\right)=\dfrac{3[\ln(x)]^2}{x}$.